Question

A hockey puck is hit on a frozen lake and starts moving with a velocity of 12.0 m/s. 5.0 seconds later, the velocity of the puck is 6.0 m/s. What is the coefficient of friction between the puck and the ice?

Answer 1

ANY MULTIPLE CHOICE

Answer 2

@riojabro

Answer 3

Can u tell me the retardation of ur puck

Answer 4

we can apply this formula:
\[\Large R\Delta t = m\Delta v\;\;\;\;\;\;\;\; (1)\]
where \(R\) is the magnitude of the friction force. So, we can write:
\[\Large R = \mu mg \;\;\;\;\;\;\;\; (2)\]
where \(\mu\) is the friction coefficient. Combining equation \(2\) with equation \(1\), we get:
\[\Large \mu mg = \frac{{m\Delta v}}{{\Delta t}}\]
therefore:
\[\Large \mu = \frac{{\Delta v}}{{g\Delta t}} = \frac{{12 - 6}}{{9.81 \cdot 5}} = ...?\]

Answer 5

Also friction force is opposite to velocity and kinetic energy
decreases. µ has no unit.